I have been studying black-body radiation and have a rudimentary understanding of how this works for solids (like a heated iron bar).

I am having trouble grasping how a "warm" gas emits a quanta of radiation.

Quoting from the Wikipedia entry,

All normal (baryonic) matter emits electromagnetic radiation when it has a temperature above absolute zero. The radiation represents a conversion of a body's internal energy into electromagnetic energy, and is therefore called thermal radiation. It is a spontaneous process of radiative distribution of entropy.

Almost all of my google queries returned discussions of greenhouse gases capturing and re-emitting IR photons due to their properties as asymmetric diatomic molecules.

That is not what I am asking. Thermal radiation from real gases (compared to an idealized black body) seems much more complicated to me due to the space between the molecules (compared to solids or liquids). It seems further complicated - for the simpler molecules that cannot vibrate like an asymmetrical diatomic molecule.

Specifically, how does a symmetrical diatomic molecule such as H2, or even a monoatomic molecule such as Ar, convert the kinetic energy of the gas molecules into electromagnetic radiation?

I would probably comprehend the answer better using a physical model. Consider one mole of H2 in a globe in space beyond our reach. The gas has a temperature of 200K and a pressure of 10 atm. The material of the globe is perfectly transparent to all radiation. The surrounding space is barren, at a temperature of 0K and there is zero background radiation.

What is the mechanism for the kinetic energy of the gas molecules to be transferred to space via radiation? How fast (approximately) would the temperature drop from 200K to 100K? How fast would the temperature drop from 5K to nearly 0K as the average kinetic energy of the molecules became very low? Would any of the answers change significantly if the gas was changed to Ar?

Thank you to anyone who chooses to answer all (or even a portion) of my oddball question!

  • 2
    $\begingroup$ You might be confused between thermal radiation and blackbody radiation. Blackbody radiation is thermal radiation, but thermal radiation is not necessarily (or even usually) blackbody radiation. $\endgroup$
    – ProfRob
    Jan 3, 2020 at 0:30
  • $\begingroup$ @RobJeffries Your answers qoudrupole moments can also been seen only another degree of freedom; movement and thus velocity. And also the emission will surely happen with some certain wavelenght; "emission sprectrum". So this anwer just adds complexity which is not needed in my opinion. But other opinions are also aloud. (at) NotAPhysicst If you change only volume of gas, there must be a simultanious change in Temperature; en.wikipedia.org/wiki/Ideal_gas_law PV=nRT where n can be seen, and R is a constant. $\endgroup$
    – Jokela
    Jan 4, 2020 at 8:13

5 Answers 5


Although electric dipolar transitions are absent for homonuclear molecules, molecular hydrogen can undergo pure rotational and vibrational-rotation transitions at temperatures of 100-1000 K in the mid-infrared (e.g. at 17 and 28 microns - see energy level diagram below, from Goldsmith et al. 2010) due to its non-zero electric quadrupole moment. In thermal equilibrium, these rotational states will be populated accordingly.

Rotational enery level diagram for hydrogen molecules from Goldsmith et al. (2010)

Lower rotational levels of the H2 molecule and quadrupole radiative transitions which connect them. For each level of para-H2, the energy in K relative to the J = 0 level is given, while for ortho-H2, the energy above the ground state of that species (J = 1) is given below the energy relative to J = 0 (in terms of a temperature $\Delta E = k_B T$). For each transition, the spontaneous decay rate in the form of the Einstein A coefficient is given below the wavelength of the transition. Taken from Goldsmith et al. (2010).

Quadrupole transitions are weak however -- the Einstein A coefficients are small. In your scenario the gas would probably cool by collisions with the walls of your containing vessel. A better question would have been to confine the gas gravitationally. This is exactly what happens during star formation in the early universe. Inefficient radiative cooling by hydrogen molecules is how the gravitational potential energy of the first collapsing protostars is radiated away (e.g. Williams 2009).

The rest of your question requires all such transitions to be identified and quantified. Your 1 mole of gas will certainly be "optically thin" to radiation, even at these transition wavelengths, so in principle, the radiative losses can be found if you know all the Einstein A coefficients and the populations of all the levels at a particular temperature. NB: You cannot use anything like the Stafan-Boltzmann law for optically thin gases.

An order of magnitude estimate would be to assume that at 200K, the $J=2-0$ transition is dominant, with an upper level population of $\sim \exp(-\Delta E/k_BT) = \exp(-510/200) = 0.078$. To work out the cooling timescale for N molecules, we equate the rate of photon emission from the excited state to the Einstein A coefficient multiplied by the number of the molecules in the excited state: $$ \frac{dN_j}{dt} \simeq A_{ji} N_j \exp(-\Delta E/k_BT)\ .$$ The cooling timescale would then be the reciprocal of the Einstein A coefficient divided by 0.078, $\sim 10^4$ years. A similar timescale arises from the $J=3-1$ transition.

This would only get you down to temperature of $\sim 100$ K. Below this, there are essentially no hydrogen molecules in the excited states and therefore no cooling via these transitions is possible. There are microwave transitions that are theoretically possible between ortho- and para- isomers, but the reciprocal Einstein coefficients for these transitions are longer than the age of the universe.

You also ask about Argon, but that is a completely different situation. Atomic argon gas has a rich infrared spectrum, that has been studied in the lab for some time (e.g. Humphreys & Kotkowski 1952), but the upper levels of these transitions aren't going to be excited by collisions at 200 K. I expect therefore that an Argon gas just couldn't cool radiatively.

  • $\begingroup$ But radiative quadrupole transitions are very weak. In practice, equilibrium is established by collisions. $\endgroup$
    – user137289
    Jan 3, 2020 at 1:23
  • $\begingroup$ @Pieter the question asked about radiative processes. So that's what my answer contains. $\endgroup$
    – ProfRob
    Jan 3, 2020 at 7:46
  • $\begingroup$ Rob, thanks for the answer! Yes, I was asking about the radiative processes only. I was trying for an example that excluded all other effects. I would have used gravitational containment - if I had even contemplated that possibility! $\endgroup$ Jan 3, 2020 at 15:01
  • 2
    $\begingroup$ My more general response is WOW! 10,000 years for the H2 to cool 50%, and Argon might not be able to radiatively cool at all. That is the part that I could not reconcile between my basic reading and the statement that, "All normal matter emits electromagnetic radiation when it has a temperature above absolute zero." $\endgroup$ Jan 3, 2020 at 15:51

Absorption is negligible in the visible and in the IR for argon and for homonuclear diatomics like H$_2$, N$_2$, and O$_2$. So according to Kirchhoff's radiation law, emissivity is negligible too. These gases do not cool by radiating.

Your gas will quickly get the temperature of the pressure vessel globe because of collisions. Then that globe was completely transparent to all radiation, so it does not radiate either. It will remain at its temperature for a very long time.

In reality, such solids do not exist. Glass will absorb and radiate at thermal wavelengths like a black body.

  • $\begingroup$ This just avoids answering the question. The OP said the container was transparent, so the gas cannot cool in this way it will just be at the same temperature as the container. Contrary to what you say, radiative cooling of hydrogen molecules is not negligible when it is all that is possible. $\endgroup$
    – ProfRob
    Jan 3, 2020 at 8:25
  • $\begingroup$ @RobJeffries In your answer you gave a quantitative estimate which confirms my handwaving "for a very long time". It is too long to wait for in a human lifetime. Negligible in comparison with other processes in a real container or when trace amounts of water etc are present. $\endgroup$
    – user137289
    Jan 3, 2020 at 9:06
  • $\begingroup$ Pieter, thanks for your answer. Yes, I did screw up my hypothetical with the container, but your answer is also enlightening. Assume the container is a perfect black body. The argon would take a very long time to cool by radiation only in my hypothetical. However, if it was in a black-body container - it would cool very quickly? The container would very efficiently radiate energy to space, and the gas would continually cool due to transferring energy via collisions with the container wall. $\endgroup$ Jan 3, 2020 at 16:03
  • $\begingroup$ @NotAPhysicist That is correct, but if the container emitted like a blackbody, you of course wouldn't be able to see the gas inside, by definition, and wouldn't have any knowledge of its properties or its radiative processes. $\endgroup$
    – ProfRob
    Jan 3, 2020 at 16:54
  • $\begingroup$ @NotAPhysicist A gas will very quickly get the same temperature as the wall of the container. A glass vessel would radiate like a black body in the IR with radiative power proportional to $T^4$ which is very slow at low $T.$ One could measure the temperature by monitoring its pressure. $\endgroup$
    – user137289
    Jan 3, 2020 at 19:37

My answer has 2 parts.

First, why do gasses emit at specific colors when heated:

The formula for black body radiation has a "hidden" assumption: The body is black, meaning it absorbs all wavelengths equally (that is actually a gray body, black also means it absorbs 100%).

This approximation is valid for a lot of solids (not all of them, for example glass) but not for gas. When a body absorbs differently different wavelengths the emitted power in these wavelengths won't obey the ratio dictated by Plank's formula. Instead, each color will have a coefficient called the emissivity which will determine how much will be emitted due to the temperature of the body being above $0^\circ$. This coefficient is deeply related to the absorption, meaning "simple atoms/molecules" (like a lot of gasses) which have characteristic absorption lines, will also emit more in these wavelengths. This gives Sodium lamps (street lamps) their yellow color, Helium lamps a different color, and so on.

Now to the second part of the answer, what is the mechanism.

This question has the same answer to the question "why does anything at all emit radiation when is above $0^\circ$?". What is black body radiation?

In thermodynamics, one can see that when in thermal equilibrium, all of the energy states accessible to the system will be excited with a certain probability (or for a macroscopic system, some part of the system is in these states). The mechanism of reaching thermal equilibrium is all of the above answers that were given before me (collisions, radiation, etc..). Meaning some atoms or molecules are in electronically excited states, and all of the time some atoms go down and some go up. But the key here is that when they go down, they sometimes emit radiation. If that radiation will get out of the system and reach our eyes, we'll see that the black body is glowing (and the system will thus lose a bit of energy). The hotter you get, the more probable your system will get to the upper energy levels, meaning they will emit higher frequencies. This is the origin of black body radiation. As you can see, the frequencies emitted are deeply connected to the energy levels, which is related to what color you absorb. So if you absorb everything you will emit everything, but if you absorb some things better than others, it means you are more probable to get to these states and emit the radiation back.

Hope I helped :)

I suggest you read more here: https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation


Q: Specifically, how does a symmetrical diatomic molecule such as H2, or even a monoatomic molecule such as Ar, convert the kinetic energy of the gas molecules into electromagnetic radiation? (What is the mechanism for the kinetic energy of the gas molecules to be transferred to space via radiation?)

A: (Short; Collision induced photon emission.) The emission and Absorbtion happens only through certain frequencies, which are known as a emission spectrum. This means that only the collisions of the particles which have the correct kinetic energy are causing this emission, and further only the radiation with a correct frequency can be absorbed. This is easy to observe in the color of bunsen burner.

This doesn't mean that gas can't cool normally. The temperature of a kinetic gas is practically it's particle velocity. Complex molecules can have various degrees of freedom, and it's only this which actually defines the heat capacity.

Q: How fast (approximately) would the temperature drop from 200K to 100K? How fast would the temperature drop from 5K to nearly 0K as the average kinetic energy of the molecules became very low?

A: Depends on boundary connections. Note that temperature (=velocity) depends also from volume.

Q: Would any of the answers change significantly if the gas was changed to Ar? A: No.

  • $\begingroup$ The paragraph with the Bunsen burner is not clear, seems confused. $\endgroup$
    – user137289
    Jan 3, 2020 at 11:59
  • $\begingroup$ Jokela, your answer states that argon would be similar to hydrogen. Rob's answer states the opposite. I do not possess the requisite physics knowledge to evaluate either position at this level. I did not follow his math for how the hydrogen quadrupole radiative transitions turn into a rate of energy loss. I assume (probably incorrectly) that even at the temperature of only 200K, a few of the gas molecules on the extreme right end of the velocity distribution reach the energy to dump a photon into space and cool the gas. $\endgroup$ Jan 3, 2020 at 16:20
  • $\begingroup$ I definitely don't understand your statement that temperature depends on volume. (I assume you mean how this is affected by boundary constraints?) If you want to expand on your answer, please remove any "complicating" conditions from my hypothetical and use any experiment that you believe is more clarifying. $\endgroup$ Jan 3, 2020 at 16:23

The Stefan-Boltzman law states $$ P= A \varepsilon\sigma T^{4}.$$ And the heat equation states $$ \frac{\partial Q}{\partial t} = c \, \rho \, \frac{\partial T}{\partial t}=K \nabla^2 T + Qgenerated$$ Assuming that the temperature is uniform we can ignore the conduction (K) term. We will also assume that the body is not absorbing any incoming radiation and that the radiation is passing through without being absorbed $$ c \, \rho \, \frac{\partial T}{\partial t} * volume= c \, m \, \frac{\partial T}{\partial t}=A \varepsilon\sigma T^{4}.$$ We will assume a mass of hydrogen of 10kilos, specific heat of 14 J/Kg, Area of 1 m^2, emissivity of 0.05, stefan boltzman constant of 5.67*10^-8 W/(K*m^2), T going from 200K to 100K and that the Temperature is only a function of the time $$ 140 \frac{dT}{dt}=2.835*10^{-9}*T^4$$ $$ 4.9*10^{10} *\frac{dT}{T^4}=dt$$ Integrate from T=200K to 100K $$ t=14291 seconds= 4hrs$$ Edited with comment suggestions

  • $\begingroup$ Your mistake is "emisivity of 1", which is incompatible with transparancy. (But this would be true for a glass pressure vessel.) $\endgroup$
    – user137289
    Jan 8, 2020 at 19:28
  • $\begingroup$ Your mistake is in assuming that the Stefan's law applies to an optically thin gas. $\endgroup$
    – ProfRob
    Jan 8, 2020 at 21:32
  • $\begingroup$ You edited to "emissivity of 0.05". But even that would be far too high for these gases. Absorption/emission of homonuclear diatomics is very low (no dipole transitions in the IR), argon even lower (no excited states at all). And it is difficult to use Stefan's law here. $\endgroup$
    – user137289
    Jan 9, 2020 at 9:18

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